Problem: Simplify the following expression: $y = \dfrac{-2x^2+7x+4}{-2x - 1}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-2)}{(4)} &=& -8 \\ {a} + {b} &=& &=& {7} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-8$ and add them together. Remember, since $-8$ is negative, one of the factors must be negative. The factors that add up to ${7}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${8}$ $ \begin{eqnarray} {ab} &=& ({-1})({8}) &=& -8 \\ {a} + {b} &=& {-1} + {8} &=& 7 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-2}x^2 {-1}x) + ({8}x +{4}) $ Factor out the common factors: $ x(-2x - 1) - 4(-2x - 1)$ Now factor out $(-2x - 1)$ $ (-2x - 1)(x - 4)$ The original expression can therefore be written: $ \dfrac{(-2x - 1)(x - 4)}{-2x - 1}$ We are dividing by $-2x - 1$ , so $-2x - 1 \neq 0$ Therefore, $x \neq -\frac{1}{2}$ This leaves us with $x - 4; x \neq -\frac{1}{2}$.